How does `cout` decide whether to print the address or the string when you pass a character pointer?

Understanding Stream Operator Overloading for Char Pointers

When printing a character pointer using printf(), the conversion specifier determines whether the address or the string is printed, such as %u for the address or %s for the string. However, with C streams and cout, how does it decide which one to output?

Consider the following code:

<code class="cpp">char ch = 'a';
char *cptr = &ch;
cout << cptr << endl;

In this example, cout attempts to interpret cptr as a string. To print the address of ch using cout, a conversion must be applied to resolve this ambiguity.

Solution: Casting to a Void Pointer

The preferred method for obtaining an address using cout is through type casting. The correct approach is:

<code class="cpp">cout << static_cast<void *>(cptr) << endl;</code>

By casting cptr to void *, we force cout to execute the appropriate overload that takes a void pointer (ostream& operator<< (ostream& o, const void *p)). This ensures that the address is printed as intended.

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